4 is the answer for your question
Answer:
σ should be adjusted at 0.5.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean 12.
Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?
13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.
So 1 should be worth two standard deviations. Then



σ should be adjusted at 0.5.
So to find a percent of increase you have to put the current number minus previous number(in this case 20-16) over the previous number(16 in this case) and multiply by 100.
So we have 20-16/16 multiplied by 100.
Since 4/16= 0.25,multiplying It by 100 gives you 25.
So the answer is 25%
mechanic #1's rate = x
mechanic #2's rate = y
* Their rate is dollars per hour ($/hr)
mechanic #1 worked for 20 hours (hr × $/hr = $)
20x = money earned by mech#1
and mechanic #2 worked for 5 hours
5y = money earned by mech#2
together they charged a total of $1150. So the amount of money earned by both mechanics.
20x + 5y = 1150
the sum of the two rates was $95 per hour.
x + y = 95
which means
x = 95 - y
plug (95 - y) in for "x" in the other equation to get everything in terms of one variable.
20(95 - y) + 5y = 1150
solve for y
1900 - 20y + 5y = 1150
1900 - 15y = 1150
-15y = 1150 - 1900
-15y = -750
y = -750/-15
y = 50 $/hr
Now use this to solve for x
x + y = 95
x + 50 = 95
x = 95 - 50
x = 45 $/hr
mech#1 charged 45$/hr
mech #2 charged 50$/hr
Answer:
0
Step-by-step explanation:
The explanation is in the attached file.
Sidenote: Hope this helps! :D